BS-6. Minimum in Rotated Sorted Array

Problem Link: bit.ly/41My2dR
Notes/C++/Java/Python codes:takeuforward.org/data-structu...
We have solved the problem, and we have gone from brute force and ended with the most optimal solution. Every approach's code has been written in the video itself. Also, we have covered the algorithm with intuition.
Full Course: bit.ly/tufA2ZYt
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0:00 Introduction of Course

[@takeUforward]

Please comment understood and give us a like if you got everything :)

[@vishnusiddarth7953]

mass

[@sudhanshushekhar4222]

Understood

[@ersoumyajitpan7205]

************ more optimized ************ int findMin(vector& nums){ int left = 0, right = nums.size() - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] > nums[right]) { left = mid + 1; } else { right = mid; } } return nums[left]; }

[@adil_k]

East or west. Striver is the best

[@aladdinstudios2735]

I came only for Minimum in Rotated Sorted Array so why I would seen previews video #striver ?

[@shivajirao999]

the feeling of writing the exact same code by your own without even looking at the lecture is insane, ALL thanks to your level of explanation

[@ravikr3463]

Abe easy hai ye upper ke 3 phele se kiye hue the esiliyee btw good

[@Rahul_Mongia]

@@ravikr3463

[@abhishekverma7604]

guys, this is one of the most important question if u r preparing for amazon and adobe.. hell they asked it a lot in previous 6 months according to Leetcode stats..

[@animeshshaw388]

Thanks brother 🙏🏾

[@akshitbhasin7322]

The fact that you acknowledged the problem of having different formats in binary search deserve a like. keep up the good work!

[@ayushpatidar5778]

The way you cover each and every edge case is amazing. Understood everything 👍

[@iWontFakeIt]

man what an explanation ! I saw several solutions on leetcode discuss but couldn't understand all of it, some if else conditions were going above my head, but you made it so clear I did code it myself both leetcode 153 and 154 with AC solutions

[@HananTariq-ws9wd]

Understood!! Hi I'm a Pakistani and I study at US. I really like your DSA playlist and recommend it to everyone. Thank you so much for your effort.

[@AmanSingh-wr6mj]

Pucha kisine?

[@vishious14]

6 minutes into the video, I was able to deduce the whole logic as soon as I understood the approach !!!!!! Amazing content !!!!!!

[@abhaymaurya9]

the same exact code i was able to write down without even watching your lecture first , kudos to u🙌❣

[@user-cx4mh7sc2j]

I'm really blessed to watch and learn the concept form you striver and i'm solving the problems now without referring the solutions. Thanks a lot.

[@shrirambalaji2915]

just watched like 7 min and understood the solution and what i am missing out thanks man you are awesome

[@surendhar.v4952]

If hurts the most , when you realize this easy question took you more than 2 hour and failed several time while in the test cases below 5 when you submit in leetcode. I did a course on java in youtube from a famous youtuber an year ago. I took a short break of two months for my semster exam and other activities. I can remind that I solved a similar type of question in an year ago. But now , I am struggling to do a easy problem like this.... Taking a break, even it is small gets your mind out of the programming mode.

[@vaibhavvm4147]

bro where did u do dsa in java from?

[@RAJSINGH-mr7hq]

Understood! Awesome explanation as always, thank you soooo much for your effort!!

[@harshitgarg2820]

Striver sir plz start a series for strings to simplify strings for us just like you did with the other topics🙏

[@snigdhadatta4702]

I was solving this for the first time , was struggling because everytime was giving me runtime error and guess what i lerally watched the first 10 mins only infinite times , because my dumb brain was just not convincing myself that how is this done, at 3pm night i passed this that too in 0ms time without seeing your written code 😭😭😭😭😭. It was giving me stress because i haven't been able to pass a single solution that day so i really needed anything to get accepted just to boost my morale . Thanks striver.

[@parth2439]

Understood, the last optimisation was great !!

[@paragroy5359]

Thanks a lot for making such videos. Your content is amazing. Keep on doing the great work

[@SahitiDantuluri]

I have gone through some solutions to this problem and able to understand but if I try to recollect and code it later I would be a little confused. But your explanation and approach is so detailed and very easy to understand that I am able to solve it on my own without any confusion (even after many days). Thanks a lot!!

[@mehulthuletiya497]

00:38 Problem statement 01:21 Brute-force approach 01:38 Optimal approach 06:30 Dry-run 10:35 Pseudocode 12:57 Code 13:20 Optimised version : If you want try to do then 15:04 Code 15:27 Complexity

[@andycharlie3255]

woow bro, this is crazy, legendary explanation, hats off man

[@Anshydv3]

understood ! the best explanation bhaiya , you are hero and a real gem❤

[@sukhjattana5887]

u unfolded the mysterious binary search....thank you!!!

[@dayashankarlakhotia4943]

Good explanation. Explanation in depth

[@Cubeone11]

I used to think that neetcode gives the best explanation, but after watching this playlist i change my thoughts.

[@MansiBansalc]

YOU ARE AN AMAZING TEACHER! THANKYOU FOR EXISTING!

[@Manasidas99]

Understood sir thank you very much, sir. Your teaching style is really amazing. I hope I will crack my interview.

[@sunnykumarpal5087]

Bhaiya your explanation is relay fabulous. It makes hard concepts easily understandable to us. Thank you bhaiya for helping us in dsa.

[@nirajaya5]

Great explanation! Thank you so much.

[@akkipinky9194]

Thanks for the incredible knowledge u give us...understood!!

[@aps2129]

Amazing explanation!!!!!

[@AnmolGupta-oj4lm]

Understood Very Well!

[@dipingrover1970]

amazing explanation . thanks a lot

[@UserUser-tn8tv]

Understood. Very Good Explanation

[@jagdishkhetre4515]

Understood...Awesome Binary search Playlist.. 👏

[@aliakbaransaria3-925]

Very good explanation Thank you

[@rahulsidhu5945]

Understood.. Awesome Binary search Playlist..😍

[@hariomtiwari9283]

Super Explanation 🎉🎉

[@namangarg8976]

Another approach I thought of - -> If left array is sorted, right is unsorted. Then go to right as pivot point can be current element or in right array -> If right is sorted, left is unsorted. Then go to left as pivot point can be current element or in left array -> If both are sorted then break and just compare the ans with arr[low] as complete search space is sorted. int findMin(vector& arr) { int low = 0; int high = arr.size() - 1; int ans = INT_MAX; while(low

[@ompandey2911]

Was wondering, What if I search for target = INT_MIN and keep the code same as the previous video wont the arr[low] be my answer?

[@myproject6768]

Absolutely understand ❤

[@rushidesai2836]

Great question!

[@visase2036]

Thanks Striver. Adding my thoughts for (duplicates) . If we apply the previous logic of high-- or low++ as the mid and low/high values are equal , we will end up getting the minimum element but that does not gaurentee the no of times array has been rotated . Examples: array=[1,1,2,1,1] , orignal array = [1,1,1,1,2]. The correct answer is 3 (as the element at 0th index has been moved to 3rd index [1,1,2,1,1]). But if we apply the previous logic, the answer would come as 0 as 0th index is the minimum element. To upsolve this, we can do the following : Keep reducing high, untill [high-1] > [high] (2>1). Once you attain this point, high(3rd index) will be the answer.

[@floatingpoint7629]

this does not cover all the cases

[@ayushmittal9666]

I think if we remove the condition of checking if(a[low]

[@md_seraj786_]

understood all clear ❤

[@tanishkthakur9965]

got it , understood everything

[@kingbadshah452]

thanks striver understood everything

[@samuelfrank1369]

Understood, Thanks a lot

[@JothiprakashThangaraj]

understood, thanks a lot!

[@NazeerBashaShaik]

Understood, thank you.

[@kiranmoura2974]

Bahut ache se smgh aaya sir ❤

[@abhaythakur2597]

very well explained

[@presidenttalks8]

Understood👍 thank you sir

[@piyushroy3278]

understood sir. Huge kudos to you :)

[@ddevarapaga5134]

Understood perfect bro

[@RishabhKumar0094]

understood, for duplicate elements worst case time complexity will be O(n/2).

[@NitinKumar-wm2dg]

understood, thank you bhaiya

[@Shunya_Advait]

Understood Sir. Thank You Sir 👌👌

[@AdityaKumar-eb9ff]

nice explanation

[@yossihadad8558]

amazing!

[@KCOYASH]

Understood :) thanks

[@UdayGarg]

bro you are amazing litt

[@motivationalcomred]

@striver take case a = [5[L] 1 2[M] 3 4[H]] mid is lesser in both arrays 2

[@infernogamer52]

Understood Bhaiya!

[@ganjinaveen7338]

Super level understood

[@mahadishakkhor123]

Hey god bless you bro❤

[@per.seus._]

UNDERSTOOD❤

[@user-or5oz1pk2x]

Thanks a lot Bhaiya

[@Rohitkumar-bx8ne]

solved without any help with BS, but before solving i have watched the previous two videos of the playlist which helped in developing the thinking skill for solving this problem

[@soumi6720]

understood!!

[@hallupandet228]

Understood💯

[@rahulreddy3588]

Understood!

[@abhay9994]

Thank you.

[@GodeDev]

Thank you

[@YourCodeVerse]

Understood✅🔥🔥

[@bilal_khan8514]

Understood Bhaiya 😃

[@rishabh1S]

Understood!!

[@senseiAree]

Understood ❤

[@shahidullahmuffakir668]

Thanks man

[@pulkitgupta669]

Understood ❤❤

[@sidusaini4658]

understood!

[@Gaurav_Tripathi_]

Understood..!!

[@heyOrca2711]

Understood! Sir

[@user-on3er8zf2n]

Understood everything

[@Aks-47]

slightly lengthier, but core logic is , we need to handle 2 cases where we are uncertain where to move, that was the crux for me, a) mid is less than lo and hi, b) mid greater than lo and hi, if we exclude that , then searching is regular BS, attaching java code for the same int ans = Integer.MAX_VALUE; while(lo

[@ANSHUKUMARI-ng4hj]

understood sir🙏

[@luvdhamija5157]

We know that if it is sorted in nature then arr[low] would always be the minimum number of all. by considering this we can simplify this as following. low=0 high=len(nums)-1 while lownums[high]: low=mid+1 elif nums[low]>nums[mid]: high=mid else: break return nums[low]

[@nihalsingh6233]

Understood!!!

[@MYMIND252]

love you bro

[@khalasianiket816]

understood❤

[@AmanKumarSharma-de7ft]

int findMin(vector& nums) { int start=0,end=nums.size()-1; while(startnums[end]?start=mid+1:end=mid;} return nums[start];} };

[@lakshminarayana5500]

Understood

[@NitinPeace]

Understood👍

[@outdoorlove-tq6qw]

i am learn from you , one day i will cross you 🙇

[@kadurkaz4691]

Love u man

[@user-cx4mh7sc2j]

Sir completed the problem which you gave as homework 😇😇

[@projectsdb4034]

understood SIR

[@baibhavghimire3827]

You beat neetcode on this🎉. Superb dude.

[@aishezsingh7004]

I think This is more intutive moving towards unsorted and if sorted move left while(low arr[mid] ) high = mid - 1; else if( arr[mid] > arr[high] ) low = mid + 1; else { ans = min(ans , arr[low] ); break; } } return ans;

[@kritikamanglam]

Understood :)

[@Santosh-up7gk]

Thank u

[@SiddharthPandey-qe2lk]

Understood🎉😮

[@pabitradas9024]

Understood bhaiya